APPLICABLE TO ELLIPTIC AND DLTEA-ELLIPTIC EUNCTIONS. 
425 
sight from the following Table, will answer the purpose ; it is a Table of the value of 
sin ‘2a: 
2a: 
COS 2a: 
and of its logarithm, from iP=45° to ^=90°. 
o 
45 
0’63662 
9-80387 
50 
0-73816 
9-86815 
55 
0-83149 
9-91986 
60 
0-91349 
9-96070 
65 
0-98041 
9-99141 
o 
70 
1-02910 
0-01246 
75 
1-05658 
0-02390 
80 
1-06216 
0-02619 
85 
1-04334 
0-01843 
90 
1 - 
0 - 
The Table shows that, past 45°, the formula 
, , , (t sin X cos x\ , 
log/=log 
is a better approximation than when the is omitted. It is to be remarked that t is 
at its minimum for x=^b°, and increases both towards and a:=90°. Near the 
latter limit, where great accuracy is required, we must proceed as follows. 
Find the correction for the logarithm of the complement of the arc by the above 
process, and then find log fi’om log y. For this purpose, I observe that 
logy=log is equivalent to y=x.l^'^\ hence 
Now, let +A=10*'— 1, whence 
log (+TOA)=log ( + ^ nearly, and also 
log(i7r-?/)=log (^- 
It is not often that the third term of either formula will be required. 
I have gone into all this detail, because the inverse tangent is continually presenting 
itself in aU these integrations, and because no book that I know shows the proper way 
of handling it. 
23. The following constants are needed for these and similar formulae : — 
10+logm =9-63778 43113 00537, 
lO+logr =8-24187 73675 90828, 
10+logF =6-46372 61172 07184, 
10+logl"=4-68557 48668 23541, 
logM=0-36221 56886 99463, 
-logr=l-75812 26324 09172, 
-log!' =3-53627 38827 92816, 
-logr=5-31442 51331 76459. 
24. As an example of finding the inverse tangent, let it be required to find log^ and 
log i^—y) from 
log tan^=9-02313 50437. Here we must take 
log tana7=9-02303 57359 
H-^= 9 93078; .-. a;=6° 1' 10"=21670" 
-^-^=302330" 
3 M 
MDCCCLXII. 
